Question: $f(x, y) = x^2 - 3xy - 1$ What are all the critical points of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(-1, 1)$ (Choice B) B $\left( 1, 0 \right)$ (Choice C) C $(0, 0)$ (Choice D) D There are no critical points.
Explanation: A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} 2x - 3y \\ \\ -3x \end{bmatrix}$ We want each component of the gradient to equal zero, so we want to solve the system of equations below. $\begin{cases} 2x - 3y = 0 \\ \\ -3x = 0 \end{cases}$ The equation $-3x = 0$ implies that $x$ must equal zero. Once we know $x = 0$, the top equation becomes $-3y = 0$, which implies that $y$ must also equal zero. There were no other cases to consider, so this is the only critical point of $f$. Therefore, $f$ has a critical point at $(0, 0)$.